Thread: itoa function trouble ??

Hi !

This is a program that i wrote to convert an integer to a string. however I came up with a problem :

The program does not work when I enter any negative number less than -999 and gives a wiered output

Can anyone help me with this and tell me why ?

I have placed ample comments in the program please let me know if still something is unclear.


The code is :

Code:

#include<stdio.h>
#include<string.h>

void itoa(int n, char s[])
{
     int i = 0;      /* variables to keep track of the index of the string */
     int j = 0;
     int k = 0;
     
     
     int sign = n;   /* variable to keep track of the sign of the integer */
     
     int temp;       /* temporary variable to store characters while reversing a string */
     
     /* make the number positive if it is negative */
     
     if (sign < 0)
        n = -n;
        
     /* run a loop and store each digit in the string ( in reverse order )*/
     
     do
     {
          s[i++] = n%10 + '0'; 
     }
     while((n/= 10) > 0);
     
     /* if the number was negative place a -ve sign at the end */
     
     if (sign < 0)
        s[i++] = '-';
     
     
     /* place a NULL character to mark the end of the string and bring the index marker i to the last character */
        
     s[i] = '\0';
     --i;
     
     /* loop to reverse the stirng */
     
    for( i = 0, j = (strlen(s) - 1); i < j; i++, j--)
     {
           temp = s[i];
           s[i] = s[j];
           s[j] = temp;
     }              
}



int main()
{
    char s[] = "";
    int n = -1000;
    itoa(n, s);
    printf("%s\n", s);
    
    return 0;
}

sorry to bother but I found my mistake

in the above program i just posted I made a mistake


" Please delete the iteration --i in line 66 "

don't need that since in the for loop I start decrementing j from (strlen(s) - 1)

sorry !!!!

Code:

 char s[] = "";

You've still got a problem because you're not allocating enough memory for the 's' char variable. This will only allocate 1 byte for the '\0'